96.4k views
5 votes
what is the standard form of the quadratic function that has a vertex of (-3,4) and a point on the graph (3,40)

User Josh Cole
by
7.2k points

1 Answer

2 votes

Final answer:

The standard form of the quadratic function with a vertex of (-3,4) and passing through the point (3,40) is y = x^2 + 6x + 13.

Step-by-step explanation:

The standard form of a quadratic function is given by y = ax^2 + bx + c. To find the constants a, b, and c, we can use the vertex form of a quadratic equation, which is y = a(x-h)^2 + k, where (h, k) is the vertex.

Since the vertex is (-3, 4), we can plug these values into the vertex form equation to get 4 = a(-3+3)^2 + 4, which simplifies to 4 = a(0)^2 + 4. This implies that the term a(0)^2 drops out and we're left with y = a(x+3)^2 + 4.

Next, we use the point on the graph (3, 40) to solve for a. Plugging in these values, we get 40 = a(3+3)^2 + 4, which reduces to 36a + 4 = 40, then 36a = 36 and finally a = 1.

With a found, we can now expand our equation y = (x+3)^2 + 4 to standard form. This gives us y = x^2 + 6x + 9 + 4, which simplifies to y = x^2 + 6x + 13.

Therefore, the standard form of the quadratic function is y = x^2 + 6x + 13.

User Gabe Morris
by
7.0k points